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Position vectors

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A,B,C be the three points in R^3 with position vectors
    a=(1,-2,0)
    b=(-1,1,2)
    c=(7,1,6)

    respectively.

    Find the position vector q of the point Q that divides the line segment AB in the ratio 2:1, (where Q is closer to B.)


    2. Relevant equations



    3. The attempt at a solution
    i have no idea how to start...
     
  2. jcsd
  3. Jun 15, 2010 #2
    do i find the length of AB first and work out the ratio?
     
  4. Jun 15, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it is not necessary to find the length of AB! You can, for instead, draw horizontal and vertical lines making a right triangle having vertices at A and B. Then "similar triangles" will make your job easy. Just divide the horizontal and vertal lines in to 3 parts.
     
  5. Jun 15, 2010 #4
    i tried to do that, but i have no idea how to find the position vector of q...
    some more hints pleasE? :p
     
  6. Jun 15, 2010 #5
    im waiting online for help, so any help is much appreciated.. thanks in advance. =)
     
  7. Jun 15, 2010 #6

    Mark44

    Staff: Mentor

    Write the vector equation for the line L that contains the segment AB. This equation will be r(t) = a + t*v

    In this equation, a is the vector from the origin to point A, and v is the vector from point A to point B. t is the parameter, and r is a vector that goes from the origin to the point on the line L determined by the parameter t.

    What vector is represented by r(0)? By r(1)? Can you think of a way to get to a point 2/3 the way along the segment AB?
     
  8. Jun 15, 2010 #7
    when r intersects the line?
     
  9. Jun 15, 2010 #8

    Mark44

    Staff: Mentor

    Each point of the line corresponds to r(t) for some value of t.
     
  10. Jun 15, 2010 #9
    so do i have to find t?
    how do i find t?
     
  11. Jun 15, 2010 #10

    Mark44

    Staff: Mentor

    You get to pick t. If I choose t = 0, r(0) = a + 0*v = <1, -2, 0>. This vector goes from the origin to point A. What I'm calling v is the vector from A to B. Since I am multiplying by 0, I don't need to do any calculations with v for this value of t.

    What is r(1)?
     
  12. Jun 17, 2010 #11
    Best way to show this is to show there exists a socalled linear combination from Linear Algebra of all three vectors which satisfies that condition.

    if the vector are called [tex]v_1,v_2,v_3[/tex]

    then a linear combination is [tex]u_1 \cdot v_1 + u_2 \cdot v_2 + u_3 \cdot v_3[/tex]

    where the [tex]u_1,u_2,u_3[/tex] are weights...
     
  13. Jun 17, 2010 #12

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since you apparently did not understand my first response, using "similar triangles" based on the coordinate axes:
    The change in x-coordinate form a(1, -2, 0) to b(-1, 1, 2) is -1- 1= -2. 2/3 of that is -4/3. That is, the change in x coordinate from a(1, -2, 0) to point Q is -4/3: the x coordinate of Q is 1+ -4/3= -1/3.

    Do the same for the y and z coordinates.
     
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